Let A be a commutative ring and S a subset of A. Let AS be the monoid of fractions of A (provided only with multiplication) with denominators in S (§ 2, no. 4). Let ε:A→AS be the canonical morphism. There exists on AS one and only one addition satisfying the following conditions: (a) AS, with this addition and its multiplication, is a commutative ring; (b) ε is a ring homomorphism.
Suppose an addition has been found for AS satisfying conditions (a) and (b).
Let x,y∈AS. Let S′ be the stable multiplicative submonoid of A generated by S. There exist a,b∈A and p,q∈S′ such that x=a/p,y=b/q. Then
x=ε(aq)ε(pq)−1,y=ε(bp)ε(pq)−1,
whence
(23)
x+y=(ε(aq)+ε(bp))ε(pq)−1=ε(aq+bp)ε(pq)−1=(aq+bp)/pq.
This proves the uniqueness of the addition.
We now define an addition on AS by setting x+y=(aq+bp)/pq. It is necessary to show that this definition does not depend on the choice of a,b,p,q. Now, if a′,b′∈A,p′,q′∈S′ are such that x=a′/p′,y=b′/q′, there exist s and t in S′ such that ap′s=a′ps,bq′t=b′qt, whence
(aq+bp)(p′q′)(st)=(a′q′+b′p′)(pq)(st)
and hence
(aq+bp)/pq=(a′q′+b′p′)/p′q′.
It is easily verified that addition in AS is associative and commutative, that 0/1 is identity element for addition, that (−a)/p is the negative of a/p and that x(y+z)=xy+xz for all x,y,z∈AS. If a,b∈A, then
ε(a+b)=(a+b)/1=a/1+b/1=ε(a)+ε(b)
and hence ε is a ring homomorphism.
Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.